MATH 4630

Chapter 3: Basic Topology of ℝ

Open Sets, Closed Sets, and Limit Points

In Chapter 2, we learned that sequences are a powerful tool for studying convergence in ℝ. But now we want to ask the question: rather than asking "does this sequence converge?", we ask "what kind of set is this?"

Question: Which subsets A ⊆ ℝ have the property that every convergent sequence (an) with an ∈ A satisfies lim an ∈ A?

Let's think about some examples:

These concepts form the foundational vocabulary of topology on ℝ, and will reappear throughout the rest of the course -- in the study of continuity, differentiation, and integration.

Open Sets:

Definition

A set O ⊆ ℝ is called open if for every point a ∈ O, there exists an ε > 0 such that the ε-neighborhood $$V_\varepsilon(a) = \underline{\hspace{5.5cm}}.$$

Intuitively, a set is open if every point has "elbow room" -- no matter where you stand inside O, you can wiggle a little in either direction and stay inside O.

Your turn: For each set, determine whether it is open using the definition. If open, exhibit an explicit ε > 0 (in terms of x) such that Vε(x) ⊆ O and verify the inclusion.
  1. O = (1,5) — Open?
  1. O = (1,5) ∪ (7,9) — Open?
  2. O = {3} — Open?
  3. O = ∅ — Open?
  4. O = [1,5) — Open?
Theorem
  1. The union of any collection of open sets is open: if {Oλ}λ∈Λ is any collection of open sets, then ⋃λ Oλ is open.
  2. The intersection of finitely many open sets is open: if O1, …, On are open, then ⋂k Ok is open.
  3. The sets ℝ and ∅ are open.

Proof of (i): Let {Oλ}λ∈Λ be any collection of open sets and let x ∈ ⋃λ Oλ.

Prove that ⋃λ Oλ is open.

Proof of (ii): Let O1, …, On be open and let x ∈ ⋂k Ok.

Show that ⋂k Ok is open.
Note: Part (ii) of the Theorem may fail for infinite intersections. Consider On = (−1/n, 1/n) for n = 1, 2, 3, …

What is ⋂n On? Is this set open?

Limit Points:

Before defining closed sets, we need a key new concept that bridges Chapter 2 and Chapter 3.

Definition

A point x ∈ ℝ is a limit point (or accumulation point) of a set A ⊆ ℝ if every ε-neighborhood Vε(x) contains a point of A different from x itself: $$\underline{\hspace{9cm}}.$$

. . . . . . . . . . . . . . . . . . . . . . . .

A point a ∈ A that is not a limit point of A is called an of A.

Note:

  1. Think of a limit point as a "popular" point -- no matter how small a neighborhood you draw around x, there is always at least one point of A nearby.
  2. Limit points do not need to be in A! -- can be inside or outside the set.
Your turn: For each set A, find all limit points and call the collection L(A).
  1. A = {1, 1/2, 1/3, 1/4, …} — L(A) =
  2. A = (0,1) — L(A) =
  3. A = ℤ — L(A) =
Theorem

A point x is a limit point of a set A if and only if there exists a sequence (an) with each an ∈ A and an ≠ x, such that (an) → x.

Proof sketch:

(⟸) Suppose (an) → x with an ∈ A, an ≠ x for all n ∈ ℕ. Let ε > 0.
Use the fact that (an) → x to show that Vε(x) contains a point of A different from x.

Therefore, x is a limit point of A. ∎

(⟹) Suppose x is a limit point of A.
For each n ∈ ℕ, since x is a limit point of A, apply the definition with ε = 1/n:

. . . . . . . . . . . . . . . . . . . . . . . .

Show that (an) → x:

Closed Sets:

Definition

A set F ⊆ ℝ is called closed if it contains all of its limit points: $$x \text{ is a limit point of } F \;\Rightarrow\; \underline{\hspace{2.3cm}}.$$

Your turn: Identify the limit points and check whether they are all inside F.
  1. F = [1,5] — Closed?
  2. F = (1,5] — Closed?
  3. F = {1/n : n ∈ ℕ} — Closed?
  4. F = {1/n : n ∈ ℕ} ∪ {0} — Closed?
  5. F = ℤ — Closed?
Theorem

A set F ⊆ ℝ is closed if and only if: whenever (an) is a convergent sequence with each an ∈ F, the limit satisfies lim an ∈ F. $$(a_n) \subseteq F \text{ and } (a_n) \to L \implies L \in F.$$

Proof sketch:

(⟹) Suppose F is closed and (an) ⊆ F with (an) → L.
Case 1: an = L for some n ∈ ℕ. Why does this immediately give L ∈ F?
Case 2: an ≠ L for all n ∈ ℕ. Use the Limit Points Theorem to conclude that L is a limit point of F. Then conclude L ∈ F:
(⟸) Suppose F is not closed. What does this mean?
That is, what is the negation of "F contains all its limit points"? Can you get a contradiction?

Open vs. Closed:

Students often assume that "open" and "closed" are opposites. They are not! A set can be both, neither, or exactly one: (0,1) is open only; [0,1] is closed only; ℝ and ∅ are both; [0,1) is neither.

Theorem

A set F ⊆ ℝ is closed if and only if its complement Fᶜ = ℝ ∖ F is open.

Proof sketch:

(⟹) Suppose F is closed and let x ∈ Fᶜ, so x ∉ F.
Since F is closed and x ∉ F, the point x is not a limit point of F. Write out explicitly what this means:
Use this to find an ε > 0 such that Vε(x) ⊆ Fᶜ, concluding that Fᶜ is open:
(⟸) Suppose Fᶜ is open. We want to show F is closed,

i.e., every limit point of F belongs to F.

Let x be a limit point of F. Suppose for contradiction that x ∈ Fᶜ. Use the openness of Fᶜ to find a neighborhood of x that misses F entirely, and explain why this contradicts x being a limit point of F:

Closed sets enjoy analogous properties to open sets, but with unions and intersections swapped.

Theorem
  1. The sets ℝ and ∅ are closed.
  2. The intersection of any collection of closed sets is closed.
  3. The union of finitely many closed sets is closed.

Proof of (ii). Let {Fλ}λ∈Λ be any collection of closed sets.

Use the Complement Theorem and De Morgan's law to show that ⋂λ Fλ is closed.
Note: Part (iii) of the Theorem may fail for infinite unions. Consider Fn = [1/n, 1] for n = 1, 2, 3, …

What is ⋃n Fn? Is this set closed?

Closure of a Set:

Definition

Let A ⊆ ℝ. The closure of A, denoted Ā, is defined as $$\overline{A} = A \cup L(A),$$ where L(A) is the set of all limit points of A.

Intuitively, Ā is the "smallest closed set containing A": we add in exactly the limit points that A is missing.

Compute the closure:
  1. A = (0,1) — Ā =
  2. A = {1/n : n ∈ ℕ} — Ā =
  3. A = ℤ — Ā =
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